Problem: Simplify the following expression: $\dfrac{8n^3}{40n^3}$ You can assume $n \neq 0$.
Answer: $ \dfrac{8n^3}{40n^3} = \dfrac{8}{40} \cdot \dfrac{n^3}{n^3} $ To simplify $\frac{8}{40}$ , find the greatest common factor (GCD) of $8$ and $40$ $8 = 2 \cdot 2 \cdot 2$ $40 = 2 \cdot 2 \cdot 2 \cdot 5$ $ \mbox{GCD}(8, 40) = 2 \cdot 2 \cdot 2 = 8 $ $ \dfrac{8}{40} \cdot \dfrac{n^3}{n^3} = \dfrac{8 \cdot 1}{8 \cdot 5} \cdot \dfrac{n^3}{n^3} $ $\phantom{ \dfrac{8}{40} \cdot \dfrac{3}{3}} = \dfrac{1}{5} \cdot \dfrac{n^3}{n^3} $ $ \dfrac{n^3}{n^3} = \dfrac{n \cdot n \cdot n}{n \cdot n \cdot n} = 1 $ $ \dfrac{1}{5} \cdot 1 = \dfrac{1}{5} $